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Calculus1
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Can someone tell me which limits i shud use for part b to find the area ?
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well you need to solve 2x - 3 = 0 that is the zero of the curve... and based on the diagram, the upper limit... the area is below the curve so remember to use absolute values
Lower limit is: x = 0 Upper limit is: x = 3/2
@aum , @campbell_st why is the lower limit zero ? why isnt it -3 ?
For any point on the y-axis, x = 0. The -3 is the y-intercept and not the x-value. The limits of integration goes from the y-axis (that is, x = 0) to where it crosses the x-axis (that is, x = 3/2).
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Oh that was so stupid of me. So thankful @aum . Aprreciate it :)
np. You are welcome.
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